Find the sum of these four vector forces: 12.0 n to the right at 35.0° above the horizontal, 31.0 n to the left at 55.0° above the horizontal, 8.40 n to the left at 35.0° below the horizontal, and 24.0 n to the right at 55.0° below the horizontal. (hint: make a drawing of this situation and select the best axes for x and y so that you have the least number of components. then add the vectors, using the component method.)

R = 7,87N to the left at 82,26° above the horizontal

Explanation:

Graph1: situation drawing (attached image)

Graph2: Resulting from the forces on the x '- y' axes (attached image)

Fx' = (12 - 8,4)N = 3,6N to the right at 35° above the horizontal

Fy' = (31 - 24)N = 7N to the left at 55° above the horizontal

Graph 3: Resulting vector of forces (R) on x-y coordinate axes (attached image)

R = Rx i + Ry j

Rx = 3,6Cos(35°) - 7Cos(55°) = -1,06N

Rx = 3,6Sin(35°) + 7Sin(55°) = 7,8N

R = (1,06 -i + 7,8 j ) N

Magnitude of R:

to the left above the horizontal.

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach